[Math] How to find the sum of the series $(1) + (2 + 3) +(4 + 5 + 6) + \cdots$ to $n$ terms

Question

How to find the sum of this series?

$$(1) + (2 + 3) +(4 + 5 + 6) + \cdots \text{ to }\, n \, \text{ terms }$$

Answer is given as:

$$\frac {1}{8} n (n + 1) (n^2 + n +2) $$

Answer 1

The $k$th term is composed of $k$ integers, so the first $n$ terms, cumulatively, will be the composed of $\displaystyle F(n) = \sum_{k=1}^{n} k = \dfrac{n(n+1)}{2}$ consecutive integers overall.

And to actually sum up those consecutive integers, we apply the same idea again.

$$S = F(F(n)) = F\left(\frac{n(n+1)}{2}\right) = \sum_{k=1}^{\frac{n(n+1)}{2}} k= \dfrac{\dfrac{n(n+1)}{2}\left(\dfrac{n(n+1)}{2}+1\right)}{2} = \frac{n (n^3 + 2 n^2 + 3 n + 2)}{8}$$

Edit: All of this relies on the fact that the sum of the first $n$ integers can be expressed in the closed form $F(n) = \dfrac{n(n+1)}{2}$ in general.

So for this problem, we're first determining how many consecutive integers show up in the first $n$ terms, which is $F(n)$. But now we need to sum the first $F(n)$ terms, so it's like computing $F(F(n))$.